All categories

2 years ago

How does 1/12 - 5/6 = 9/12 um tring to learn how to add this like ?

Mathematics

2 answers:

tekilochka [14]2 years ago

6 0

Answer:

see below

Step-by-step explanation:

1/12 - 5/6 = -9/12

We need to get a common denominator of 12

1/12 - 5/6*2/2 = -9 /12

1/12 - 10/12 = -9/12

Since the denominators are the same, we can subtract the numerators

1-10 = -9

-9/12 = - 9/12

We can simplify -9/12 by dividing the top and bottom by 3

-9/3 = -3

12 /3 = 4

-9/12 = -3/4

harina [27]2 years ago

3 0

Answer:

see below

Step-by-step explanation:

1/12 - 5/6 = -9/12

1st, we need to identify the Least Common Multiplier of 12 and 6

in this case, the LCM is 12

for 5/6 = (5*2) / (6*2) = 10/12

2nd, adjust the fractions based on the LCM

= 1 / 12 - 10 / 12

= 1 - 10 / 12

= -9 / 12 ---- simplify -3 / 4

You might be interested in

How would you find the area of this parallelogram? Describe your strategy.

fiasKO [112]

Answer:

To find the area of a parallelogram, multiply the base by the height.

The formula is: A = B x H where B is the base, H is the height.

Step-by-step explanation:

6 0

2 years ago

Use the functions f(x) and g(x) to answer the question. F(x)=x²-16; g(x)=x+4

AnnZ [28]

(x^2) - 16 / x+4 = (x-4)(x+4) / (x+4) = x-4, x not equal to -4

D is the answer.

7 0

3 years ago

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .